Direction of gradient from level surface? In the diagram below, we see a level surface with a gradient.
As a consequence of the multivariable chain rule, the gradient is normal to the surface. That's clear to me.
Why is the gradient pointing outward rather than into the sphere?
I understand that if it were, it would be a negative gradient, but that's a consequence, not an explanation.
Why does the gradient have to point away from a level set rather than into it?

 A: Gradient points towards the direction in which function value is increasing (in maximum sense). I guess the value of the function is increasing as you move away from the origin as I can see from the level curves. Hence it is pointing away from the origin.
Elaboration:
Say you are in mountain, then all rings at constant heights are level curves. Now the gradient is the direction in which if you move the value of the function (the height) will increase, also if you take the negative gradient direction, you will start descending. Question is why it happens.
The answer lies in how slope is defined. Take one variable case, say $\frac{\partial f}{\partial x} $, if this is positive then going along the direction of the slope (i.e. increasing $x$) will increase the value of my function. If the slope is negative, the function value will start increasing as we move along the slope (in -ive $x$ direction, remember the slope is negative) . The similar analogy applies to multivariable case. 
Please refer a good text (may be Thomas), first learn how slope is defined in arbitrary direction, and then see how it applies in gradient case (using dot product).
A: That's not always true.
In general, the gradient points towards arguments in which function value is higher. Therefore, a gradient pointing "outwards" indicates that we need to move outwards to find higher function values, but sometimes we find higher function values by moving inwards. This depends on what "outwards" means, because we usually get it from context and don't define it rigorously. And it also depends on which function we use to define the gradient, because we don't get a gradient from a surface alone.
Consider the sphere $x^2+y^2+z^2=r^2$. This is a level set of $f(x,y,z)=x^2+y^2+z^2$, and in this case the gradient of $f$ points "outwards" because we took $f$ such that higher values of $f(x,y,z)$ correspond to spheres with higher radii. But we could also consider this sphere as a level set of $g(x,y,z)=(x^2+y^2+z^2)^{-1}$ or $h(x,y,z)=-(x^2+y^2+z^2)$ and in this case the gradient points inwards, because higher values of $g(x,y,z)$ correspond to spheres with smaller radii.
Similarly, the level curves of the mountain $f(x,y)=-x^2-y^2$ are circumferences and for this $f$ the gradient points inwards. But for $g(x,y)=x^2+y^2$ (the valley) or $h(x,y)=(-x^2-y^2)^{-1}$ the gradient points outwards.
Of course, we could get a different surface normal by taking the opposite of one of these gradients or by redefining the gradient in such a way that it points in the direction of decrease, but that's not the question here. We're assuming the usual definition of gradient, that is coherent with the way derivatives are defined in the one variable case.
